How do you rewrite $\sqrt{3} sin θ + cos θ$ $sqrt(3)sin theta + costheta$ as a sum?

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Answer 1 · 2016-11-23T14:29:35

We know that ${sin}^{2} θ + {cos}^{2} θ = 1$ $sin^2theta + cos^2theta = 1$ . Isolating ${cos}^{2} θ$ $cos^2theta$ , we have:

${cos}^{2} θ = 1 - {sin}^{2} θ$ $cos^2theta = 1 - sin^2theta$

$cos θ = \sqrt{1 - {sin}^{2} θ}$ $costheta = sqrt(1 - sin^2theta)$

$\Rightarrow \sqrt{3} sin θ + \sqrt{1 - {sin}^{2} θ}$ $=>sqrt(3)sin theta + sqrt(1 - sin^2theta)$

Hopefully this helps!

Answer 2 · 2016-11-23T16:03:05

Given expression

$= \sqrt{3} sin θ + cos θ$ $=sqrt3sintheta+costheta$

$\Rightarrow 2 (\frac{\sqrt{3}}{2} sin θ + \frac{1}{2} cos θ)$ $=>2(sqrt3/2sintheta+1/2costheta)$

$\Rightarrow 2 (cos (\frac{π}{6}) sin θ + sin (\frac{π}{6}) cos θ)$ $=>2(cos(pi/6)sintheta+sin(pi/6)costheta)$

$\Rightarrow 2 (sin θ cos (\frac{π}{6}) + cos θ sin (\frac{π}{6}))$ $=>2(sinthetacos(pi/6)+costhetasin(pi/6))$

$\Rightarrow 2 sin (θ + \frac{π}{6})$ $=>2sin(theta+pi/6)$
(applying formula $sin A cos B + cos A sin B = sin (A + B)$ $sinAcosB+cosAsinB=sin(A+B)$

How do you rewrite √3sinθ+cosθsqrt(3)sin theta + costheta as a sum?